A-Stability and Stiff Problems Explained Visually | Numerical Methods – Lecture 42

Some differential equations have perfectly smooth solutions, yet explicit methods such as Euler and RK4 crawl through them with tiny steps or blow up. For y′ = −1000(y − cos t) the solution hugs cos t, but Euler needs h below 0.002: with h = 0.0021 it explodes near t = 0.42. These are stiff problems: a fast decaying component, not accuracy, dictates the step of an explicit method. The cure is backward Euler, yₙ₊₁ = yₙ + h f(tₙ₊₁, yₙ₊₁), an implicit method solved by Newton's method at each step. Its stability function G(z) = 1/(1 − z) is stable outside the unit disk around 1, so it is A-stable: stable for every h whenever Re λ ≤ 0. Explicit methods have polynomial G and bounded regions, so they can never be A-stable. We work the notes' examples (y′ = t + y, y′ = −50y with h = 0.05, y′ = −y³), compare with the trapezoidal rule, count the steps, and close the course with a look back over all its modules. 📌 Chapters 0:00 Introduction 0:33 Recap: stability regions 1:27 A smooth solution 1:58 Two time scales 2:26 Euler blows up 2:59 The stability limit 3:31 What is a stiff problem? 4:08 Systems: the eigenvalues decide 4:54 Backward Euler: the idea 5:19 An implicit formula 5:50 Solving each step with Newton 6:34 Worked example: y′ = t + y 7:23 The test equation 7:51 The stability region 8:21 A-stability 8:46 Explicit methods are never A-stable 9:28 y′ = −50y, h = 0.05 10:21 Zigzag against decay 10:49 The stiff problem solved 11:32 The trade: cost per step 12:01 A nonlinear implicit step 12:55 The trapezoidal rule 13:50 Algorithm 14:29 Comparison 15:15 Summary 16:00 Looking back over the course 17:02 Thank you 📝 Key results • A-stable: the whole left half-plane Re z ≤ 0 lies in the stability region • Backward Euler: yₙ₊₁ = yₙ + h f(tₙ₊₁, yₙ₊₁), G(z) = 1/(1 − z), stable for |1 − z| ≥ 1, A-stable, order 1 • Explicit Runge–Kutta methods have polynomial G and bounded regions: never A-stable • Stiff: a fast decaying component forces explicit methods to tiny steps (h ≤ 2/|λ| for Euler) • y′ = −1000(y − cos t): Euler needs h ≤ 0.002 (over 5000 steps to t = 10); backward Euler with h = 0.1 takes 100 steps, error about 5 × 10⁻⁵ • y′ = −50y, h = 0.05: Euler multiplies by −1.5 each step, backward Euler by 1/3.5 ≈ 0.2857 • Each implicit step solves g(Y) = Y − yₙ − h f(tₙ₊₁, Y) = 0, by Newton's method 💻 Practise this lecture Self-paced practice with live plots, hints and instant feedback: 👉 https://biswarupbiswas.github.io/nume... 📚 Course website (all lectures, notes and practice): https://biswarupbiswas.github.io/nume... This is the last lecture of the course. Thank you for watching: every lecture, with notes and practice, stays on the course website. #NumericalMethods #StiffEquations #DifferentialEquations #ODE #Mathematics #Engineering